BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

(x^8y^2+x^2y^8)/(x^2+y^2)=?

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members
Location: Las Vegas, USA
Thanked: 19 times
by Max@Math Revolution » Mon Dec 31, 2018 1:13 am

Timer

00:00

Answers

A

0%

B

0%

C

0%

D

0%

E

0%

Stats

Difficulty—

0% CorrectAvg —

0% IncorrectAvg —

[Math Revolution GMAT math practice question]

(x^8y^2+x^2y^8)/(x^2+y^2)=?

A. x^2y^2(x^4-x^2y^2+y^4)
B. x^2y(x^4+x^2y^2+y^4)
C. x^2y^2(x^2+y^2)
D. x^2y^2(x+y)
E. x^4y^4(x^2-y^2)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members
Location: Las Vegas, USA
Thanked: 19 times

by Max@Math Revolution » Wed Jan 02, 2019 12:26 am
=>

(x^8y^2+x^2y^8)/(x^2+y^2)= x^2y^2(x^6+y^6)/(x^2+y^2)= x^2y^2(x^2+y^2)(x^4-x^2y^2+y^4)/(x^2+y^2)= x^2y^2 (x^4-x^2y^2+y^4), using the identity a^3+b^3 = (a+b)(a^2-ab+b^2).

Therefore, the answer is A.
Answer: A
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members
Thanked: 59 times

by fskilnik@GMATH » Wed Jan 02, 2019 2:03 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

(x^8y^2+x^2y^8)/(x^2+y^2)=?

A. x^2y^2(x^4-x^2y^2+y^4)
B. x^2y(x^4+x^2y^2+y^4)
C. x^2y^2(x^2+y^2)
D. x^2y^2(x+y)
E. x^4y^4(x^2-y^2)
Let´s explore a particular case and proceed with the "filtering" (i.e., refuting alternatives that do not match the target)!
$$? = {{{x^8}{y^2} + {x^2}{y^8}} \over {{x^2} + {y^2}}}\,\,$$
$${\left. {{{{x^8}{y^2} + {x^2}{y^8}} \over {{x^2} + {y^2}}}\,\,} \right|_{\,\left( {x,y} \right) = \left( {1,1} \right)}}\, = \,\,\,{{1 + 1} \over {1 + 1}}\,\,\, = 1\,\,\,\,\,\,\,\left( {{\rm{TARGET}}} \right)$$
$$\left. \matrix{
\left( {\rm{A}} \right)\,\,\,1 \cdot \left( {1 - 1 + 1} \right) = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{survivor}}\,\, \hfill \cr
\left( {\rm{B}} \right)\,\,\,1 \cdot \left( {1 + 1 + 1} \right) \ne 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refuted}} \hfill \cr
\left( {\rm{C}} \right)\,\,\,1 \cdot \left( {1 + 1} \right) \ne 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refuted}} \hfill \cr
\left( {\rm{D}} \right)\,\,\,1 \cdot \left( {1 + 1} \right) \ne 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refuted}} \hfill \cr
\left( {\rm{E}} \right)\,\,\,1 \cdot \left( {1 - 1} \right) \ne 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{refuted}} \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{A}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1